Chapter 04 Rotational Motion

P. J. Grandinetti

Chem. 4300

P. J. Grandinetti Chapter 04: Rotational Motion Angular Momentum Angular momentum of particle with respect to origin, O, is given by

퓁⃗ = ⃗r × p⃗ Rate of change of angular momentum is given z by cross product of ⃗r with applied force.

p m d퓁⃗ dp⃗ = ⃗r × = ⃗r × F⃗ = ⃗휏 r dt dt

O y Cross product is defined as applied torque, ⃗휏.

x

Unlike linear momentum, angular momentum depends on origin choice.

P. J. Grandinetti Chapter 04: Rotational Motion Conservation of Angular Momentum Consider system of N Particles

z

m5 m 2 Rate of change of angular momentum is

m3 ⃗ ∑N 퓁⃗ ∑N ⃗ m1 dL d 훼 dp훼 = = ⃗r훼 × dt dt dt 훼=1 훼=1 y which becomes

m4 x ⃗ ∑N dL ⃗ net = ⃗r훼 × F dt 훼 Total angular momentum is 훼=1

∑N ∑N ⃗ ⃗ L = 퓁훼 = ⃗r훼 × p⃗훼 훼=1 훼=1

P. J. Grandinetti Chapter 04: Rotational Motion Conservation of Angular Momentum

⃗ ∑N dL ⃗ net = ⃗r훼 × F dt 훼 훼=1 Taking an earlier expression for a system of particles from chapter 1 ∑N ⃗ net ⃗ ext ⃗ F훼 = F훼 + f훼훽 훽=1 훽≠훼 we obtain ⃗ ∑N ∑N ∑N dL ⃗ ext ⃗ = ⃗r훼 × F + ⃗r훼 × f훼훽 dt 훼 훼=1 훼=1 훽=1 훽≠훼 and then obtain 0 > ⃗ ∑N ∑N ∑N dL ⃗ ext ⃗ rd ⃗ ⃗ = ⃗r훼 × F + ⃗r훼 × f훼훽 double sum disappears from Newton’s 3 law (f = −f ) dt 훼  12 21 훼=1 훼=1 훽=1 훽≠훼 P. J. Grandinetti  Chapter 04: Rotational Motion Conservation of Angular Momentum

⃗ ∑N ∑N dL ⃗ ext = ⃗r훼 × F = ⃗휏훼 = ⃗휏 dt 훼 total 훼=1 훼=1

If there is no net external torque on system of particles then system’s total angular momentum, L⃗, is constant,

dL⃗ if ⃗휏 = 0, then = 0, and L⃗ = constant total dt This is the principle of conservation of angular momentum.

True in quantum mechanics as well as in classical mechanics.

P. J. Grandinetti Chapter 04: Rotational Motion Orbital and Spin Angular Momentum

P. J. Grandinetti Chapter 04: Rotational Motion Orbital and Spin Angular Momentum Consider system of particles again. Total angular momentum relative to origin is ∑N ∑N ⃗ ⃗ ⃗ dr훼 L = 퓁 = ⃗r훼 × m훼 i dt 훼=1 훼=1 COM Defining position of each particle relative to center of mass, ⃗ ⃗r훼 = R + r⃗◦훼

⃗r훼 is position of particle position relative to origin, R⃗ is center of mass relative to origin, O r⃗◦훼 is particle position relative to center of mass. Expression for L⃗ becomes (see notes for derivation) ∑N ⃗ ⃗ ⃗ dr◦훼 L = R × p⃗ + r⃗◦훼 × m훼 ⏟⏞⏞⏟⏞⏞⏟total dt 훼=1 relative to origin ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ relative to CM P. J. Grandinetti Chapter 04: Rotational Motion Orbital and Spin Angular Momentum

∑N ⃗ ⃗ ⃗ dr◦훼 L = R × p⃗ + r⃗◦ × m훼 total 훼 dt ⏟⏞⏞⏟⏞⏞⏟ 훼=1 relative to origin ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ relative to CM

Identify total angular momentum as split into orbital and spin angular momentum terms: ⃗ ⃗ ⃗ L = Lorbital + Lspin

⃗ Imagine Lorbital as angular momentum of Earth as it moves around sun as the origin, ⃗ and Lspin as angular momentum of Earth as it spins about its center of mass. Often true, to a good approximation, that orbital and spin parts are separately conserved.

P. J. Grandinetti Chapter 04: Rotational Motion Rotational energy

Total kinetic energy of system of particles is ( ) ∑N 2 1 d⃗r훼 K = m훼 dt 훼=1 2 ⃗ Written in terms of R and r⃗◦훼 gives ( ) ( ) ⃗ 2 ∑N 2 1 dR 1 dr⃗◦훼 K = M + m훼 2 dt 2 훼 dt ⏟⏞⏞⏞⏞⏟⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟=1 center of mass rotation about center of mass 1st term is energy associated with motion of center of mass of system.

2nd term is energy associated with rotational motion of system about center of mass.

P. J. Grandinetti Chapter 04: Rotational Motion Rigid Bodies

P. J. Grandinetti Chapter 04: Rotational Motion Rigid Bodies

Definition When all particles in system are rigidly connected we have a rigid body. This ideal model assumes no relative movement of composite particles.

z m5 m2

m3

m1 y

m4 x

P. J. Grandinetti Chapter 04: Rotational Motion How do we specify the orientation of a rigid body?

Definition Euler’s rotation theorem states that orientation of rigid body in a given coordinate system can be described by rotation through an angle about a single axis.

Only 2 angles needed to define orientation of rotation axis so full orientation of any rigid body about fixed point can be described by just 3 parameters: e.g., polar, 휃, and azimuthal, 휙, angles and angle of rotation, 휒:

Sign of ⃗휒 is determined by right-hand rule: thumb points along rotation axis and right-hand fingers curl in direction of positive rotation. Set of 휙, 휃, and 휒 are called Euler angles

P. J. Grandinetti Chapter 04: Rotational Motion Euler angles Another example of a set of Euler angles is convention for giving orientation of an airplane by 3 parameters called yaw, pitch, and roll:

Pitch Axis (y)

Roll Axis Yaw Axis (x) (z)

Yaw, pitch, and roll are just one of many ways of defining the 3 angles implied by Euler’s rotation theorem.

P. J. Grandinetti Chapter 04: Rotational Motion Angular velocity vector, ⃗휔

Definition Angular velocity vector, ⃗휔, is vector passing through origin along axis of rotation and whose magnitude equals magnitude of angular velocity,

⃗휔 휔 ⃗ = er 휔 휒 ⃗ = d ∕dt and er is unit vector defined by ⃗ 휃 휙 ⃗ 휃 휙 ⃗ 휃 ⃗ er = sin cos ex + sin sin ey + cos ez

Direction of ⃗휔 is determined by right-hand rule: Curl right-hand fingers in direction of rotation, then thumb points in direction of ⃗휔. Keep in mind when describing motion of rigid body that magnitude and orientation of ⃗휔 can change with time.

P. J. Grandinetti Chapter 04: Rotational Motion Angular velocity vector, ⃗휔 Best origin choice through which ⃗휔 passes depends on rigid body motion being described. For molecular rotations : natural origin choice is molecule’s center of mass. For top spinning on table surface : origin is better located at fixed point where tip of top meets table surface.

z y

x

y

x

P. J. Grandinetti Chapter 04: Rotational Motion Relationship between angular momentum and angular velocity vectors Total angular momentum of rigid body relative to origin is ∑N ∑N ⃗ J = (⃗r훼 × m훼v⃗훼) = m훼(⃗r훼 × v⃗훼) 훼=1 훼=1 Note: notation change, J⃗ instead of L⃗ for rigid body. th Linear velocity vector, v⃗훼, of 훼 particle is related to its angular velocity, ⃗휔, by

v⃗훼 = ⃗휔 × ⃗r훼

⃗r훼 is particle position. ⃗휔 is identical for all particles in rotating rigid body since all inter-particle distances are constant. Combining these two expressions ... ∑N [ ] ⃗ J = m훼 ⃗r훼 × ( ⃗휔 × ⃗r훼) 훼=1 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ vector triple product

P. J. Grandinetti Chapter 04: Rotational Motion Relationship between angular momentum and angular velocity vectors Total angular momentum of rigid body relative to the origin is

∑N [ ] ⃗ J = m훼 ⃗r훼 × ( ⃗휔 × ⃗r훼) 훼=1 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ vector triple product

Vector triple product has a well known expansion ( ) a⃗ × b⃗ × c⃗ = (a⃗ ⋅ c⃗) b⃗ − (a⃗ ⋅ b⃗) c⃗

With triple product expansion we get

∑N [( ) ( ) ] ⃗ J = m훼 ⃗r훼 ⋅ ⃗r훼 ⃗휔 − ⃗r훼 ⋅ ⃗휔 ⃗r훼 훼=1

P. J. Grandinetti Chapter 04: Rotational Motion Relationship between angular momentum and angular velocity vectors Moment of Inertia Tensor

∑N [( ) ( ) ] ⃗ J = m훼 ⃗r훼 ⋅ ⃗r훼 ⃗휔 − ⃗r훼 ⋅ ⃗휔 ⃗r훼 훼=1 can be rewritten as matrix equation

⎛ ⎞ ⎛ ⎞ ⎛ 휔 ⎞ Jx Ixx Ixy Ixz x ⃗ ⎜ ⎟ ⎜ ⎟ ⎜ 휔 ⎟ ⋅ ⃗휔 J = ⎜ Jy ⎟ = ⎜ Iyx Iyy Iyz ⎟ ⎜ y ⎟ = I 휔 ⎝ Jz ⎠ ⎝ Izx Izy Izz ⎠ ⎝ z ⎠ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏟⏟ I ⃗휔

matrix I is called the moment of inertia tensor (see next slide).

J⃗ = I ⋅ ⃗휔 When rigid body rotates about origin with angular velocity ⃗휔 the size and direction of body’s angular momentum, J⃗, is determined by its moment of inertia tensor, I.

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor of a rigid body Elements of Moment of Inertia Tensor about the center of mass If rigid body rotates about its center of mass then relevant moment of inertia tensor–relative to center of mass—is calculated according to ∑N ( ) ∑N 2 2 2 2 Ixx = m훼 y훼 + z훼 − M(Y + Z ) Ixy = Iyx = − m훼x훼y훼 + MXY 훼=1 훼=1 ∑N ( ) ∑N 2 2 2 2 Iyy = m훼 x훼 + z훼 − M(X + Z ) Iyz = Izy = − m훼y훼z훼 + MYZ 훼=1 훼=1 ∑N ( ) ∑N 2 2 2 2 Izz = m훼 x훼 + y훼 − M(X + Y ) Ixz = Izx = − m훼x훼z훼 + MXZ 훼=1 훼=1 M is total mass X, Y, Z are coordinates of the center of mass.

Notice that off-diagonal elements are symmetric about diagonal, Ixy = Iyx, Ixz = Izx, etc. Tensors with this property are called symmetric tensors.

P. J. Grandinetti Chapter 04: Rotational Motion The Principal Axis System of a Moment of Inertia Tensor Off-diagonal elements are symmetric about the diagonal

⎛ Ixx Ixy Ixz ⎞ ⎜ ⎟ I = ⎜ Ixy Iyy Iyz ⎟ ⎝ Ixz Iyz Izz ⎠ For any symmetric tensor one can always find an axis system in which it is diagonal.

⎛ Ixx Ixy Ixz ⎞ ⎛ Ia 0 0 ⎞ PAS 휙, 휃, 휒 ⋅ ⎜ ⎟ ⋅ T 휙, 휃, 휒 ⎜ ⎟ I = R( ) ⎜ Ixy Iyy Iyz ⎟ R ( ) = ⎜ 0 Ib 0 ⎟ ⎝ Ixz Iyz Izz ⎠ ⎝ 0 0 Ic ⎠

Diagonal elements, Ia, Ib, and Ic, called the principal moments of inertia. R(휙, 휃, 휒) is a rotation matrix and RT (휙, 휃, 휒) is its transpose. Definition Principal Axis System (PAS) is a coordinate system in which I is diagonal. ≥ ≥ Convention is to assign axes of body-fixed frame, a, b, and c, such that Ic Ib Ia.

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor Principal Axis System

When moment of inertia tensor is calculated in the PAS we have

⎛ Ia 0 0 ⎞ PAS ⎜ ⎟ I = ⎜ 0 Ib 0 ⎟ ⎝ 0 0 Ic ⎠

Symmetry can help identify the principal axis system. A 2-fold or higher symmetry axis in a molecule is a principal axis.

If 3 principal moments of inertia are equal then any axis in space can be a principal axis.

If 2 principal moments are equal then any axis in plane of the 2 equal principal moments can be a principal axis.

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor Principal Axes

For freely rotating object, i.e.,with no external torque applied, J⃗ is conserved, dJ⃗∕dt = 0.

If freely rotating object has constant (i.e., time independent) angular velocity vector, ⃗휔, then ⃗휔 points along a principal axis, and J⃗ is aligned with ⃗휔 (recall J⃗ = I ⋅ ⃗휔).

If ⃗휔 of freely rotating object is not aligned with J⃗, then ⃗휔 is time dependent since conservation of angular momentum requires J⃗ to remain time-independent.

When viewed (calculated) in body-fixed frame I is time-independent.

When viewed (calculated) in space-fixed frame I may be time dependent as rigid body rotates.

P. J. Grandinetti Chapter 04: Rotational Motion Intermediate Axis Theorem aka Tennis Racket Theorem aka the Dzhanibekov1 effect

Definition ≠ ≠ Intermediate axis theorem states that the rotation of an object with Ia Ib Ic around its Ia and Ic principal axes is stable, while rotation around the Ib axis (or intermediate axis) is not.

Ic

Dzhanibekov effect

Intermediate Axis Theorem Ia

Ib

1Soviet Cosmonaut P. J. Grandinetti Chapter 04: Rotational Motion P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u (1 u = 1.66053904 × 10−27 kg).

z

# atom mass x y z 1 H 1 u +0.5826 Å 0 +0.7592 Å . 105° x 2 H 1 u +0.5826 Å 0 −0 7592 Å 3 O 16 u 0 0 0

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u (1 u = 1.66053904 × 10−27 kg).

We know M = 18 u, and X = 0.06473 Å, and Y = Z = 0. Diagonal moment of inertia components are

∑N 0 :0 2¡ 2 2 2 Ixx = m훼(y¡훼 + z훼) − M(Y + Z ) # atom mass x y z 훼=1 1 H 1 u +0.5826 Å 0 +0.7592 Å ∑N ( ) 0 2 2 2 2 2 H 1 u +0.5826 Å 0 −0.7592 Å Iyy = m훼 x훼 + z훼 − M(X + Z ) 훼 3 O 16 u 0 0 0 =1 ∑N 0 0 2 2¡ 2 2 Izz = m훼(x훼 + y¡훼 ) − M(X + Y ) 훼=1

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u (1 u = 1.66053904 × 10−27 kg).

We know M = 18 u, and X = 0.06473 Å, and Y = Z = 0. Diagonal moment of inertia components are

2 2 u . Å 2 u . Å 2 . −47m2⋅kg Ixx = mHz1 + mHz2 = (1 )(0 7592 ) + (1 )(−0 7592 ) = 1 914 × 10 ( ) ( ) 2 2 2 2 2 Iyy = mH x(1 + z1 + mH x2 + z2 − MX) ( ) . 2 . 2 . 2 . 2 . 2 Iyy = (1 u) (0 5826 Å) + (0 7592Å) + (1 u) (0 5826 Å) + (−0 7592Å) − (18 u)(0 06473Å) . −47 2⋅ Iyy = 2 916 × 10 m kg

2 2 2 Izz = mHx1 + mHx2 − MX . 2 . 2 . 2 . −47 2⋅ Izz = (1 u)(0 5826 Å) + (1 u)(0 5826 Å) − (18 u)(0 06473Å) = 1 002 × 10 m kg

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u.

And the off-diagonal components are

Ixy = Iyx = 0

Iyz = Izy = 0

Ixz = Izx = −mHx1z1 − mHx2z2 . . . . Ixz = Izx = −(1 u)(0 5826 Å)(0 7592Å) − (1 u)(0 5826 Å)(−0 7592Å) = 0

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule. Take OH bond length as d = 0.957 Å, H-O-H angle as Ω = 105◦, and O and H masses as 16.0 u and 1.0 u.

In the coordinate system for this calculation we obtain a diagonal moment of inertia tensor given by

⎛ 1.914 0 0 ⎞ −47 2⋅ ⎜ . ⎟ I∕10 m kg = ⎜ 0 2 916 0 ⎟ ⎝ 0 0 1.002 ⎠

Not done yet! Must follow convention for labeling the axes in principal axis system. In this > > example we identify (Iyy = Ic) (Ixx = Ib) (Izz = Ia).

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule.

Final answer Water molecule has moment of inertia tensor in the PAS given by

z

⎛ 1.002 0 0 ⎞ PAS −47 2⋅ ⎜ . ⎟ I ∕10 m kg = ⎜ 0 1 914 0 ⎟ ⎝ 0 0 2.916 ⎠ 105° x

When labeling axes we need to maintain right-handed coordinate system where a⃗ × b⃗ = c⃗.

P. J. Grandinetti Chapter 04: Rotational Motion Moment of Inertia Tensor

Example Calculate moment of inertia tensor of water molecule.

Final answer Water molecule has moment of inertia tensor in the PAS given by

⎛ 1.002 0 0 ⎞ PAS −47 2⋅ ⎜ . ⎟ I ∕10 m kg = ⎜ 0 1 914 0 ⎟ 105° b ⎝ 0 0 2.916 ⎠

a When labeling axes we need to maintain right-handed coordinate system where a⃗ × b⃗ = c⃗.

P. J. Grandinetti Chapter 04: Rotational Motion Rotational Energy of Rigid Bodies and Classification of molecules

P. J. Grandinetti Chapter 04: Rotational Motion Rotational energy of a rigid body th In PAS kinetic energy only comes from rotational motion. Define v⃗◦훼 as velocity of 훼 particle relative to center of mass. ∑N ∑N 1 2 1 K = m훼v⃗◦ = m훼v⃗◦훼 ⋅ v⃗◦훼 2 훼 2 훼=1 훼=1

Since v⃗◦훼 = ⃗휔 × r⃗◦훼 we can write ∑N ∑N ∑N 1 1 1 K = m훼v⃗◦훼 ⋅ ( ⃗휔 × r⃗◦훼) = ⃗휔 ⋅ m훼(r⃗◦훼 × v⃗◦훼) = ⃗휔 ⋅ r⃗◦훼 × p⃗훼 2 2 2 훼=1 훼=1 훼=1 ∑ ⃗ N ⃗ ⃗ Recognizing final sum as total angular momentum, J = 훼=1 r◦훼 × p훼, we have 1 K = ⃗휔 ⋅ J⃗ 2 Since J⃗ = I ⋅ ⃗휔 we obtain our expression for the kinetic energy of the rotational motion of a rigid body, 1 K = ⃗휔 ⋅ I ⋅ ⃗휔 2

P. J. Grandinetti Chapter 04: Rotational Motion Rotational energy of a rigid body

1 K = ⃗휔 ⋅ I ⋅ ⃗휔 2 Kinetic energy reduces to convenient form in PAS (i.e., a body-fixed frame) in terms of principal moments: 1 ( ) K = 휔2I + 휔2I + 휔2I 2 a a b b c c

In PAS we also have 휔 , 휔 , 휔 Ja = Ia a Jb = Ib b and Jc = Ic c which we use to rewrite kinetic energy as

J2 J2 J2 K = a + b + c 2Ia 2Ib 2Ic

P. J. Grandinetti Chapter 04: Rotational Motion Classification of molecules

Molecules grouped into 5 classes based on principal components.

Name Diagonal values Examples Spherical Ia = Ib = Ic = I CH4 < Prolate Symmetric I|| = Ia Ib = Ic = I⟂ CH3F < Oblate Symmetric I⟂ = Ia = Ib Ic = I|| CHF3 < < Asymmetric Ia Ib Ic CH2Cl2, CH2CHCl Linear Ia = 0, Ib = Ic = I OCS, CO2

Molecules with two or more 3-fold or higher rotational symmetry axes are spherical tops. All molecules with one 3-fold or higher rotational symmetry axis are symmetric tops because principal moments about two axes normal to n-fold rotational symmetry axis (n ≥ 3) are equal.

P. J. Grandinetti Chapter 04: Rotational Motion Spherical: Ia = Ib = Ic = I Methane is an example of a molecule with this rotational symmetry. H

. −40 ⋅ 2 H C H Ia = Ib = Ic = 5 3 × 10 g cm

H Spherical molecules have more than one 3-fold axis of symmetry and all 3 principal moments of inertia are equal. Principal moments of inertia of spherical tetrahedral and octahedral molecules:

In spherical symmetry case kinetic energy takes form

2 2 2 2 Ja + J + Jc J K = b = 2I 2I

P. J. Grandinetti Chapter 04: Rotational Motion Symmetric molecules: two principal moments are equal

Symmetric molecules have a three-fold or higher axis of symmetry.

Presence of this symmetry axis leads to two of the principal moments of inertia being equal.

3rd principal moment is associated with rotation about axis with highest rotational symmetry, called the figure axis.

In prolate and oblate symmetric tops the figure axis is associated with the smallest and largest principal moment of inertia, respectively.

P. J. Grandinetti Chapter 04: Rotational Motion < Prolate Symmetric: I|| = Ia Ib = Ic = I⟂ think “cigar-shaped”

Label perpendicular moment of inertia component I⟂ for Ib = Ic and parallel component I|| for Ia.

Fluoromethane is an example of a molecule with prolate symmetry. H . −40 ⋅ 2 → Ia = 5 3 × 10 g cm F C H a . −40 ⋅ 2 Ib = Ic = 32 9 × 10 g cm H

In prolate symmetric case kinetic energy expression takes form ( ) J2 J2 + J2 2 a b c J 1 1 2 K = + = + − Ja 2I|| 2I⟂ 2I⟂ 2I|| 2I⟂

P. J. Grandinetti Chapter 04: Rotational Motion < Oblate Symmetric: I⟂ = Ia = Ib Ic = I|| think “frisbee-shaped”

Label perpendicular moment of inertia component I⟂ for Ia = Ib and parallel component I|| for Ic.

Trifluoromethane is an example of a molecule with oblate symmetry. F

. −40 ⋅ 2 → Ia = Ib = 81 1 × 10 g cm H C F c . −40 ⋅ 2 Ic = 149 1 × 10 g cm

F

In oblate symmetric case kinetic energy expression takes form ( ) J2 + J2 J2 2 a b c J 1 1 2 K = + = + − Jc 2I⟂ 2I|| 2I⟂ 2I|| 2I⟂

P. J. Grandinetti Chapter 04: Rotational Motion Symmetric Rotor Generic Cases

P. J. Grandinetti Chapter 04: Rotational Motion < < Asymmetric: Ia Ib Ic < < Molecules with Ia Ib Ic have lowest symmetry.

Dichloromethane is example of molecule with asymmetric symmetry. H Cl . −40 ⋅ 2 a Ia = 26 41 × 10 g cm ↑ . −40 ⋅ 2 C Ib = 254 51 × 10 g cm → . −40 ⋅ 2 b Ic = 275 71 × 10 g cm H Cl Asymmetric molecules can possess two-fold axes and planes of symmetry but cannot have any three-fold or higher symmetry axes.

In asymmetric case kinetic energy expression cannot be simplified from general form

J2 J2 J2 K = a + b + c 2Ia 2Ib 2Ic

P. J. Grandinetti Chapter 04: Rotational Motion Linear: Ia = 0, Ib = Ic = I

Linear molecules have an axis with infinite rotational symmetry (C∞) since all the atoms lie on one axis.

Carbon dioxide and OCS are two examples of a molecule with linear symmetry. , → Ia = 0 O C S a . −40 ⋅ 2 Ib = Ic = I = 138 0 × 10 g cm , → Ia = 0 O C O a . −40 ⋅ 2 Ib = Ic = I = 71 3 × 10 g cm

P. J. Grandinetti Chapter 04: Rotational Motion Linear Molecule: General Expressions

P. J. Grandinetti Chapter 04: Rotational Motion Linear Molecule: Kinetic Energy To obtain kinetic energy expression in linear case we need to reconsider derivation.

Because Ia = 0 there can be no rotational motion around the line of atoms. All rotation occurs about an axis perpendicular to the a axis. 휔 Since Ia = 0 and a = 0 we modify 1 ( ) K = 휔2I + 휔2I + 휔2I 2 a a b b c c to be 1 ( ) 1 K = 휔2 + 휔2 I = 휔2I 2 b c 2 2 ⃗ ⋅ ⃗ 2휔2 2휔2 2휔2 Similarly we see that J = J J = Ib b + Ic c = I and we have

J2 K = 2I as kinetic energy of linear rigid rotor.

Warning: looks the same as spherical case, but not the same. Linear case derived with Ia = 0.

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor

Previously, we worked in body-fixed coordinate systems with axes chosen along molecular axes of highest rotational symmetry where I is diagonal.

If different coordinate system was used for water:

z we would have calculated

⎛ 1.576 0 0.441 ⎞ I∕10−47m2⋅kg = ⎜ 0 2.916 0 ⎟ 105° ⎜ ⎟ ⎝ 0.441 0 1.340 ⎠ x

How do we find principal components of I and transformation back to coordinate system where I is diagonal?

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor When angular velocity vector, ⃗휔, is aligned with a principal axis we have

⎛ ⎞ ⎛ ⎞ Ia 0 0 0 ⃗ ⃗휔 ⎜ ⎟ ⎜ ⎟ ⃗휔 J = I c = ⎜ 0 Ib 0 ⎟ ⎜ 0 ⎟ = Ic c 휔 ⎝ 0 0 Ic ⎠ ⎝ ⎠ ⃗휔 In this case, c is vector aligned with a principal axis with principal component value Ic. We can write 3 equations:

⃗휔 ⃗휔 , ⃗휔 ⃗휔 , ⃗휔 ⃗휔 I a = Ia a I b = Ib b I c = Ic c

| ⃗휔 | Divide both sides by corresponding magnitude i , and rearrange to ( ) ⃗ I − Ii1 ei = 0 ⃗ ⃗휔 ei is unit vector along direction of i, and 1 is identity matrix. This is a set of 3 simultaneous linear equations for each principal component.

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor ( ) ⃗ I − Ii1 ei = 0

1 To find principal components we expand determinant | | | I − I I I | ( ) | xx i xy xz | | | det I − Ii1 = | Iyx Iyy − Ii Iyz | = 0 | | | Izx Izy Izz − Ii |

to get a 3rd-order polynomial equation in Ii. The 3 principal components Ia, Ib, Ic are roots of this 3rd-order polynomial equation

2 PAS orientation obtained by substituting each principal component back into top equation ( ) ( ) ( ) ⃗ , ⃗ , ⃗ , I − Ia1 ea = 0 I − Ib1 eb = 0 I − Ic1 ec = 0 ⃗ and solving for corresponding ei which points along direction for Ii in original coordinate system where I was non-diagonal.

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor Example Given the inertia tensor calculated in an arbitrary body-fixed frame shown below

⎛ 1.576 0 0.441 ⎞ −47 2⋅ ⎜ . ⎟ I∕10 m kg = ⎜ 0 2 916 0 ⎟ ⎝ 0.441 0 1.340 ⎠

determine (a) its principal components of the inertia tensor and (b) the orientation of the PAS.

(a) For the general form of this tensor in this problem we can create the determinant, | | | Ixx − Ii 0 Ixz | | | | 0 I − I 0 | = 0 | yy i | | Ixz 0 Izz − Ii | and expand to obtain the polynomial

2 2 (Ii − (Ixx + Izz)Ii + IxxIzz − Ixz)(Iyy − Ii) = 0

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor Finding principal components from roots of polynomial

2 2 (Ii − (Ixx + Izz)Ii + IxxIzz − Ixz)(Iyy − Ii) = 0 . −47 2⋅ As expected, one root is I1 = Iyy = 2 916 × 10 m kg, leaving us to find other 2 roots from

2 2 Ii − (Ixx + Izz)Ii + IxxIzz − Ixz = 0

Applying quadratic formula we find 2 solutions [ √ ] 1 I = I + I + (I − I )2 + 4I2 = 1.914 × 10−47m2⋅kg 2 2 xx zz xx zz xz

and [ √ ] 1 I = I + I − (I − I )2 + 4I2 = 1.002 × 10−47m2⋅kg 3 2 xx zz xx zz xz , < , < Finally, follow convention and assign Ia = I3 Ib = I2 Ic = I1

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor Finding principal axis system , < , < (a) Following convention assign Ia = I3 Ib = I2 Ic = I1, and obtain

⎛ 1.002 0 0 ⎞ −47 2⋅ ⎜ . ⎟ I∕10 m kg = ⎜ 0 1 914 0 ⎟ ⎝ 0 0 2.916 ⎠

(b) To find orientation of each principal axis substitute principal component values back into ( ) ( ) ( ) ⃗ , ⃗ , ⃗ , I − Ia1 ea = 0 I − Ib1 eb = 0 I − Ic1 ec = 0

Obtain 3 simultaneous equations for each principal axis direction.

These simultaneous equations are usually redundant and only relationships among xi, yi, and zi are obtained, rather than unique values for them.

Normalization of unit vectors gives additional constraint to pin down values for xi, yi, and zi.

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor

Finding the principal axis system : I1 case . −47 2⋅ Starting with I1 = Ic = 2 916 × 10 m kg, we obtain ⎛ . . . ⎞ ⎛ ⎞ ⎛ . . ⎞ ⎛ ⎞ 1 576 − 2 916 0 0 441 xc −1 34 0 0 441 xc ⎜ . . ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ 0 2 916 − 2 916 0 ⎟ ⎜ yc ⎟ = ⎜ 0 0 0 ⎟ ⎜ yc ⎟ = 0 . . . . . ⎝ 0 441 0 1 340 − 2 916 ⎠ ⎝ zc ⎠ ⎝ 0 441 0 −1 576 ⎠ ⎝ zc ⎠

Obtain two equations: . −1 34xc + zc = 0 . . 0 441xc − 1 576zc = 0

Satisfy two equations with xc = 0, zc = 0 while letting yc take on any possible value. In other words, solution to these equations is any point on line along y axis. ⃗ ⃗ Define unit vector along direction associated with Ic as ec = ey

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor

Finding the principal axis system : I2 case . −47 2⋅ Switching to I2 = Ib = 1 914 × 10 m kg, we obtain ⎛ . . . ⎞ ⎛ ⎞ ⎛ . . ⎞ ⎛ ⎞ 1 576 − 1 914 0 0 441 xb −0 338 0 0 441 xb ⎜ . . ⎟ ⎜ ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ 0 2 916 − 1 914 0 ⎟ ⎜ yb ⎟ = ⎜ 0 1 002 0 ⎟ ⎜ yb ⎟ = 0 . . . . . ⎝ 0 441 0 1 340 − 1 914 ⎠ ⎝ zb ⎠ ⎝ 0 441 0 −0 574 ⎠ ⎝ zb ⎠

. . −0 338xb + 0 441zb = 0 . 1 002yb = 0 . . 0 441xb − 0 574zb = 0 . Satisfy equations by setting yb = 0, xb = 1 and solving for zb = 0 76644. From this solution we make vector e⃗ + 0.76644 e⃗ pointing in direction associated with I . √ x z b 2 2 . Using xb + zb = 1 256 we normalize vector to make unit vector pointing in Ib direction 1 ( ) e⃗ = e⃗ + 0.76644 e⃗ = 0.7937 e⃗ + 0.6083 e⃗ b 1.256 x z x z

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor

Finding the principal axis system : I3 case . −47 2⋅ Finally for I3 = Ia = 1 002 × 10 m kg, we obtain ⎛ . . . ⎞ ⎛ ⎞ ⎛ . . ⎞ ⎛ ⎞ 1 576 − 1 002 0 0 441 xa 0 574 0 0 441 xa ⎜ . . ⎟ ⎜ ⎟ ⎜ . ⎟ ⎜ ⎟ ⎜ 0 2 916 − 1 002 0 ⎟ ⎜ ya ⎟ = ⎜ 0 1 914 0 ⎟ ⎜ ya ⎟ = 0 . . . . . ⎝ 0 441 0 1 340 − 1 002 ⎠ ⎝ za ⎠ ⎝ 0 441 0 0 338 ⎠ ⎝ za ⎠

. . 0 574xa + 0 441za = 0 , 1 914ya = 0 . . 0 441xa + 0 338za = 0 . Satisfy equations by setting ya = 0, xa = 1 and solving for za = −1 30159. From this solution make vector e⃗ − 1.30159 e⃗ pointing in direction associated with I . √ x z a 2 2 . Using xa + za = 1 6414 normalize vector to make unit vector pointing in Ia direction 1 ( ) e⃗ = e⃗ − 1.30159e⃗ = 0.6092e⃗ − 0.7930e⃗ a 1.6414 x z x z

P. J. Grandinetti Chapter 04: Rotational Motion Diagonalizing a moment of inertia tensor Finding the principal axis system ⃗ ⃗ Directions of ea and eb vectors are draw in figure below.

z

⃗ . ⃗ . ⃗ ea = 0 6092ex − 0 7930ez 105° ⃗ . ⃗ . ⃗ eb = 0 7937 ex + 0 6083 ez x ⃗ ⃗ ec = −ey

⃗ ⃗ ⃗ Note: sign change in ec = −ey needed to obtain right-handed coordinate system. Direction of ec vector is directed out of and perpendicular to plane of page. Consistent with our earlier moment of inertia calculation on water.

P. J. Grandinetti Chapter 04: Rotational Motion Rotating frame transformation

P. J. Grandinetti Chapter 04: Rotational Motion Rotating frame transformation

When rigid body rotates the “body-fixed” frame becomes a non-inertial frame.

Newton’s laws suggest that they would no longer apply in such a non-inertial (i.e., accelerating or rotating) frame.

There are approaches to obtain correct equations of motion in rotating frame which can be related back to some inertial “space-fixed” frame.

P. J. Grandinetti Chapter 04: Rotational Motion Rotating frame transformation In inertial “space-fixed” frame a vector Q⃗ can be decomposed into its projections onto unit vectors ⃗ ⃗ ⃗ ⃗ Q = Qxex + Qyey + Qzez Rate of change of Q⃗ in inertial “space-fixed” frame is ( ) ⃗ dQ dQ dQy dQ = x e⃗ + e⃗ + z e⃗ dt dt x dt y dt z space Q⃗ can also be can be decomposed into its projections onto unit vectors in a non-inertial rotating frame ⃗ ⃗ ⃗ ⃗ Q = Qaea + Qbeb + Qcec Rate of change of Q⃗ in rotating frame is ( ) dQ⃗ dQ dQ dQ = a e⃗ + b e⃗ + c e⃗ dt dt a dt b dt c rot ( ) ( ) How are dQ⃗ ∕dt and dQ⃗ ∕dt related? space rot P. J. Grandinetti Chapter 04: Rotational Motion Rotating frame transformation Calculate rate of change of Q⃗ in space-fixed frame in terms of its decomposition in rotating frame. ( ) dQ⃗ dQ dQ dQ de⃗ de⃗ de⃗ = a e⃗ + b e⃗ + c e⃗ + Q a + Q b + Q c dt dt a dt b dt c a dt b dt c dt space ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ( ) dQ⃗ ∕dt rot ( ) We identify first 3 terms on right as dQ⃗ ∕dt rot ⃗ ⃗ ⃗ Unit vectors in rotating frame, ea, eb, and ec, are time dependent in space-fixed frame from rotation, so last 3 derivatives are non-zero.

Linear velocity of unit vector as it rotates about a given direction with angular velocity, Ω⃗ , is given by cross product ( ) de⃗ ⃗ i ⃗ ⃗ vi = = Ω × ei dt space

P. J. Grandinetti Chapter 04: Rotational Motion Rotating frame transformation

Last term can be rewritten de⃗ de⃗ de⃗ Q a + Q b + Q c = Q Ω⃗ × e⃗ + Q Ω⃗ × e⃗ + Q Ω⃗ × e⃗ = Ω×⃗ Q e⃗ + Ω×⃗ Q e⃗ + Ω×⃗ Q e⃗ = Ω⃗ × Q⃗ a dt b dt c dt a a b b c c a a b b c c

Leaving us with ( ) ( ) dQ⃗ dQ⃗ = + Ω⃗ × Q⃗ dt dt space rot

This relates rate of change of Q⃗ in space-fixed frame to its rate of change in frame rotating with Ω⃗ .

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion

Relationship between rate of change of J⃗ in space- and body-fixed frames is ( ) ( ) dJ⃗ dJ⃗ = + ⃗휔 × J⃗ dt space dt body

In inertial space-fixed frame ( ) dJ⃗ = ⃗휏 dt space

Combining top 2 equations gives ( ) dJ⃗ + ⃗휔 × J⃗ = ⃗휏 dt body

This is Euler’s equation of motion for a rigid body.

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion ( ) dJ⃗ + ⃗휔 × J⃗ = ⃗휏 dt body Working in body-fixed frame we relate J⃗ and ⃗휔 by

⃗ 휔 ⃗ 휔 ⃗ 휔 ⃗ J = aIaea + bIbeb + cIcec

Substituting this into Euler’s equations of motion (top of slide) and get 3 coupled differential equations: ( ) d휔 a 휔 휔 휏 Ia − (Ib − Ic) b c = a ( dt )body d휔 b 휔 휔 휏 Ib − (Ic − Ia) c a = b ( dt )body d휔 c 휔 휔 휏 Ic − (Ia − Ib) a b = c dt body

What happens if there is no external torques (⃗휏 = 0), that is, free rotation?

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion – Free rotation When there is no external torque, that is, ⃗휏 = 0, Euler’s equations describe free rotation of rigid body ( ) ( ) ( ) d휔 (I − I ) d휔 (I − I ) d휔 (I − I ) a b c 휔 휔 , b c a 휔 휔 , c a b 휔 휔 = b c = c a = a b dt body Ia dt body Ib dt body Ic

What happens when there’s no external torque and rigid body is rotating about a principal axis? ≠ ≠ ⃗ 휔 휔 If rigid body with Ia Ib Ic is initially rotating about a principal axis, say ec, then a = b = 0. ( ) ( ) ( ) d휔 d휔 d휔 a = b = c = 0 dt body dt body dt body Rate of change of all 3 components of ⃗휔 is zero: → magnitude and orientation of ⃗휔 remains constant in rigid body frame. ⃗ 휔 ⃗ Since J = Ic cec: → magnitude and orientation of J⃗ remains constant in rigid body frame and aligned with ⃗휔. Since J⃗ is conserved for freely rotating rigid body (i.e., ⃗휏 = 0): → magnitude and orientation of J⃗ remains remains constant in space-fixed inertial frame, and ⃗휔 is aligned with J⃗ P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion When there is no external torque, that is, ⃗휏 = 0, Euler’s equations describe free rotation of rigid body ( ) ( ) ( ) d휔 (I − I ) d휔 (I − I ) d휔 (I − I ) a b c 휔 휔 , b c a 휔 휔 , c a b 휔 휔 = b c = c a = a b dt body Ia dt body Ib dt body Ic

What happens when there’s no external torque and rigid body is rotating about an arbitrary axis?

⃗휔 becomes time dependent in body-fixed frame.

J⃗ becomes time dependent in body-fixed frame since J⃗ = I ⋅ ⃗휔. Remember, I is time-independent in body-fixed frame.

Since J⃗ is conserved for freely rotating rigid body (i.e., ⃗휏 = 0): → J⃗ is time independent in space-fixed inertial frame. → ⃗휔 is NOT time independent in space-fixed frame. Remember, I may be time dependent in space-fixed frame.

P. J. Grandinetti Chapter 04: Rotational Motion Free rotation of oblate symmetric rigid body about an arbitrary axis < where Ia = Ib = I⟂ Ic = I||

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion < Free rotation of oblate symmetric rigid body (Ia = Ib = I⟂ Ic = I||) about an arbitrary axis ( ) ( ) ( ) d휔 (I − I ) d휔 (I − I ) d휔 (I − I ) a b c 휔 휔 , b c a 휔 휔 , c a b 휔 휔 = b c = c a = a b dt body Ia dt body Ib dt body Ic < In case of oblate symmetric body, where Ia = Ib = I⟂ Ic = I||, Euler’s equations simplify to (I − I ) (I − I ) ̇휔 ⟂ || 휔 휔 ̇휔 ⟂ || 휔 휔 ̇휔 a = c b b = − c a c = 0 I⟂ I⟂

Using Newton’s dot notation for time derivatives. Define new constant angular frequency, Ω0, as I − I ⟂ || 휔 Ω0 = c I⟂ and further simplify two coupled differential equations to: ̇휔 휔 ̇휔 휔 a = Ω0 b b = −Ω0 a

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion < Free rotation of oblate symmetric rigid body (Ia = Ib = I⟂ Ic = I||) about an arbitrary axis

̇휔 휔 ̇휔 휔 a = Ω0 b b = −Ω0 a

To solve these coupled differential equations we take time derivative again to obtain

̈휔 ̇휔 ̈휔 ̇휔 a = Ω0 b and b = −Ω0 a

Substitute 1st-derivative expressions into 2nd-derivative expressions to get 2 uncoupled 2nd-order homogeneous differential equations:

̈휔 2휔 ̈휔 2휔 a + Ω0 a = 0 and b + Ω0 b = 0

For each equation we can propose a solution of the form

휔 i(t) = Ai cos kit + Bi sin kit

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis Substituting 휔 i(t) = Ai cos kit + Bi sin kit into ̈휔 2휔 ̈휔 2휔 a + Ω0 a = 0 and b + Ω0 b = 0 gives 2 2 2 2 Ai(Ω0 − ki ) cos kit + Bi(Ω0 − ki ) sin kit = 0

To make equation true for all values of t we set ki = Ω0 and obtain 휔 휔 a(t) = Aa cos Ω0t + Ba sin Ω0t and b(t) = Ab cos Ω0t + Bb sin Ω0t 휔 휔 휔 Choose initial condition: a(t = 0) = ⟂ and b(t = 0) = 0 leads to 휔 휔 Aa = ⟂, Ba = 0, and Ab = 0, Bb = ⟂, and finally to 휔 휔 휔 휔 a(t) = ⟂ cos Ω0t and b(t) = ⟂ sin Ω0t ⃗휔 휔 Projection of onto a-b plane has constant length, ⟂, and rotates in a-b plane with frequency of Ω0. P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis

휔 ⃗휔 Since c is constant then total length of is constant, √ 휔 휔2 휔2 휔2 = a + b + c

⃗휔 ⃗ Define angle makes with ec with 훼 휔 휔 tan = ⟂∕ c and get [ ] ⃗휔 휔 훼 ⃗ 훼 ⃗ 훼 ⃗ (t) = sin cos Ω0t ea − sin sin Ω0t eb + cos ec

⃗휔 ⃗ In body-fixed frame precesses about ec with body-fixed frame 훼 angle of at frequency Ω0.

P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis Motion of J⃗ in body-fixed frame is [ ] ⃗ ⋅ ⃗휔 휔 훼 ⃗ 훼 ⃗ 훼 ⃗ J = I = I⟂ sin cos Ω0t ea − I⟂ sin sin Ω0t eb + I|| cos ec

⃗ ⃗ J has constant length and precesses around ec at angular frequency Ω0.

⃗ ⃗ Angle between J and ec is

J I⟂ tan 휃 = a = tan 훼 Jc I||

In oblate case I⟂ < I|| and 휃 is smaller than 훼 ⃗휔 ⃗ ⃗ As both and J rotate around ec all 3 vectors remain in same plane. body-fixed frame P. J. Grandinetti Chapter 04: Rotational Motion Euler’s equations of motion Free rotation of oblate symmetric rigid body about an arbitrary axis

⃗ ⃗휔 ⃗ ⃗ In space-fixed frame we know that J remains unchanged, thus and ec must precess about J.

body-fixed frame space-fixed frame

P. J. Grandinetti Chapter 04: Rotational Motion